Frame Flex on GCN

[Mark McM]

Okay, so here's a quick summary of what's been discussed: I think we can all agree to the following:

The frame deflects under vertical pedal forces;

The primary deflection mode of the frame is a torsional rotation of the BB shell around the longitudinal axis;

The torsional rotation of the BB shell causes the crankarms/pedals to rise or fall, depending on which direction of rotational flex;

As the forces on the pedals continue to change during the crank rotation, the frame un-flexes, on its way to reversing its direction of flex;

Energy stored in frame flex is not lost, but is restored in some way to the system when the frame unflexes;

So the key question is, just where does this energy go?

It appears that the answer to this question is largely dependant on where in the crank rotation the frame unflexes, which is itself dependant on how the pedal forces vary during the crank rotation. That sounds simple, but in reality, there can be a lot of variation in how the pedal forces vary during crank rotation.

The amount of torsion flex at the BB will be dependent on the right/left pedal force differential. In the pedal force vector diagram below, the right/left pedal differential is maximized around 6 o'clock. This means the downward flex at a pedal increases throughout the downstroke, and then decreases on the upstroke. So the flexural energy is restored as the pedal rises, acting to lift the rearward leg. None of the restored energy goes directly into the drivetrain, but instead helps the rider to move their legs.



But in this second pedal force vector example just below, the right/left pedal differential (and therefore the maximal downward flex at the pedal) is maximized around 4 o'clock. This means that much of the unflexing occurs while the downward moving crank is still forward of the BB. The unflexing of the frame acts to raise the BB while the pedal is still descending. This results in forward rotation of the crank. In this case, some of the energy stored in the frame goes directly into generating crank torque, and therefore directly into the drivetrain. (Note: most of the frame unflexing in this case still occurs in a portion of the crank rotation where the 'spring' force/restored energy primarily acts to raise the bottom leg).



In both cases, it is a case of a zero sum gain. The energy stored in the frame is resturned, and helps to drive the bike forward either indirectly or directly: Indirectly, in that it reduces the amount of energy the rider needs to expend to bring their leg back up for the next power stroke; or directly, in that the energy goes into the drivetrain during the power stroke. Or, most likely in many cases, a little of both.

[William]

Originally Posted by Kontact

I honestly can't tell if his frame is flexing or if his wheels and fork are. I can't see the HT and ST going out of plane with each other.

If you look at the faceplate on the stem when he first starts torquing it (before the camera pulls right) it looks like the head tube and seat tube are moving out of plane with each other. How much is frame and how much is wheel/fork is hard to tell for sure.





William

[Kontact]

Originally Posted by Mark McM

Well, that's clearly not true. The frame is flexing precisely because we are stepping on it (or rather, because we are stepping on the pedal). If we step on the pedal when the crank is forward of the BB, then we generate a torque on the crank to drive the front wheel. But if we step on the pedal when the crank is vertical, no drive torque is generated, but we still flex the frame just the same.

And yet, the BB flex goes away before we pedal through 6 where you've shown that there is still plenty of down force on the pedal - down force that is much more in line with BB flex than it even is at 3 o'clock.

Maybe you think about this way, but you're thinking is incorrect. You keep claiming that the stays get shorter and longer, but unfortunately there is no evidence that they do to any meaningful extant. So there can be virtually no energy storage due to this type of deflection. The deflection modes of the frame were modelled and analyzed in a web page that was reference on the 3rd page of this discussion (see link below). This analysis shows that the primary energy storage in the frame is torsional deflection around the the longitudinal axis. In other words, the cranks/pedals are going up and down due to a torsional twisting of the frame, causing a rotation of the BB shell as the frame twists.

What is that, a royal fiat?


Did you read the article you linked to?

For a final part of this analysis, I have used an FEA model to determine the response of a frame to the four loads described earlier. It is somewhat trivial to look at how much energy goes in and out of the frame since we know that it gets released into the drive train. But it is interesting to understand about how much energy gets temporarily stored in the frame.
....

Having concluded that frame flex does not waste energy, I do not believe that frame stiffness is irrelevant.

I have offered the "twisting stays" model as a basic way of thinking about what happens when you have a plane between two lines (chain stays between hub and BB shell), and you move one of those points. I'm not talking about where the energy is specifically stored, but how that energy is redelivered to the drivetrain. Obviously, the energy is stored in every component that is bending - especially the down tube, seat tube and chain stays.

You might not like that way of looking at it, but frankly the ability to model the actual geometry of all of this is a bit much for anyone to do in their head, so we have to talk about this with simplified versions. So here's a very simple example of the problem:

If I take a flat strip of spring steel and twist the ends in opposite directions, would you not agree that the overall length of the strip gets shorter?

When I release that twist, does the strip not push itself back to full length?

And if I cut a slot down the center of the strip like the space between the chainstays, does any of that change?

[kramnnim]

So Kirk is saying that the flexed frame, returning back to neutral as the rider is still applying pressure, is directing the energy to the rear wheel... If the frame moves 5mm at the peak of the pedal stroke, and was able to flex back those 5mm while the rider was still applying pressure, the crank arm would rotate slightly. (5mm?)

That does make sense...as long as the frame is flexing back against the pedal stroke. If it can't flex back until 6 oclock, the energy would not go back through the chain to the rear wheel and would be wasted.

[Kontact]

Originally Posted by William

If you look at the faceplate on the stem when he first starts torquing it (before the camera pulls right) it looks like the head tube and seat tube are moving out of plane with each other. How much is frame and how much is wheel/fork is hard to tell for sure.

Not saying it isn't happening, I just can't tell if what I'm looking at is twist or just what happens when something leans out of plane and one part is closer to the observer than the others. There is a perspective issue with vanishing points that makes it very hard to observe if the head tube and seat tube aren't actually crossing in view.

[Kontact]

Originally Posted by Mark McM

None of the restored energy goes directly into the drivetrain, but instead helps the rider to move their legs.

This is one of those statements like "the man's feet don't push against the earth, the earth pushed up on his feet."

You can't argue that "helping the rider move their legs" is different than "helping the rider move the crank" in a system where net force never goes to zero. Until force goes to zero, the bike and the legs are all one mechanical linkage and you don't get to pick where the force is acting within it.

[Kontact]

Originally Posted by kramnnim

That does make sense...as long as the frame is flexing back against the pedal stroke. If it can't flex back until 6 oclock, the energy would not go back through the chain to the rear wheel and would be wasted.

Why not? Did the the rider input go to zero and the system went completely slack?

Under normal pedaling do you ever feel the freewheel disengage, even momentarily, at any part of the pedal stroke?

[cachagua]

The bike and the legs are all one mechanical linkage and you don't get to pick where the force is acting within it.

No, we don't get to choose. You're right about that.

I still think the force will act where it is resisted less. It sure would put my mind at rest if someone could say how it would not do that, but instead would act on the thing that's resisting it more...?

[Kontact]

Originally Posted by cachagua

No, we don't get to choose. You're right about that.

I still think the force will act where it is resisted less. It sure would put my mind at rest if someone could say how it would not do that, but instead would act on the thing that's resisting it more...?

It's a fair question. I found this rather scholarly article that is more about how to get results than about the results themselves, but it does show that pedaling is always net positive torque value. See figure 6 on page 7:

https://www.uni-konstanz.de/mmsp/pub...s/QuDaSa15.pdf

It might be useful to know just how much torque it takes to notably flex a frame, because that number might compare usefully to the minimum pedaling torque to suggest how much un-flexing could feed back against pedaling. It would also be good to know when the unflexing happens - at minimum torque or somewhere before or after it.

[cachagua]

It's a fair question. I found this rather scholarly article that is more about how to get results than about the results themselves, but it does show that pedaling is always net positive torque value...

I don't quite see the connection with what I was asking. Or is that point related to something else?

It would also be good to know when the unflexing happens - at minimum torque or somewhere before or after it.

Is "when it happens" really the way to think about it? It certainly doesn't occur all at once -- we could say it begins to happen when your power decreases below its maximum, and continues to happen more as you decrease more -- and then, on into the next stroke when you begin to push harder again, the frame begins to flex again (in the direction appropriate to that stroke, of course). Insofar as the flex is a response to the power you're putting out, it will vary in tandem with that.

And clarifying things this far is making me begin to wonder if the BB swaying Mark describes as a result of non-torque pedal pressure may have to be dealt with separately. Can we conveniently separate frame flex that's a result of torque, and frame flex that's a result of non-torque forces?

[kramnnim]

Originally Posted by Kontact

Why not? Did the the rider input go to zero and the system went completely slack?

Under normal pedaling do you ever feel the freewheel disengage, even momentarily, at any part of the pedal stroke?

Indoors at low power and low RPM, sitting up, hand off the bars...yes, I hear several clicks of the freewheel at the 12/6 dead spot.

No, it isn't normal pedaling, but it exaggerates just how dead the dead spot is.

If we compare the motion of the BB flexing back against the pedal stroke to when you are up out of the saddle, rocking the bike back and forth, using your arms to rock the bike left as your right foot pushes down...that movement does nothing useful at 12/6.

That type of rocking would be one of the times when a frame flexes the most, and it seems like the flexed frame would be rebounding right around 11/5 to 12/6. So it would be wasted just like bobbing around on a full suspension MTB. Or are you going to say that the energy stored in the compressed suspension is also pushed back into the drivetrain...

[Kontact]

Originally Posted by cachagua

I don't quite see the connection with what I was asking. Or is that point related to something else?



Is "when it happens" really the way to think about it? It certainly doesn't occur all at once -- we could say it begins to happen when your power decreases below its maximum, and continues to happen more as you decrease more -- and then, on into the next stroke when you begin to push harder again, the frame begins to flex again (in the direction appropriate to that stroke, of course). Insofar as the flex is a response to the power you're putting out, it will vary in tandem with that.

And clarifying things this far is making me begin to wonder if the BB swaying Mark describes as a result of non-torque pedal pressure may have to be dealt with separately. Can we conveniently separate frame flex that's a result of torque, and frame flex that's a result of non-torque forces?

The point I was making is that torque never comes off the chain and that torque has a minimum level. Despite that minimum level the BB flex comes out, in part because it is pulled to the other side by the alternating pedaling rather than just the drop in pedal torque. So when the BB starts to move back to center and then gets to center, how much tension was on the chain, when was it on it and how much of BB recentering is just decreases in torque and how much is the other leg doing to tug it that way?

[cachagua]

When the BB... gets to center, how much tension was on the chain?

Less than some other place in the pedal stroke. That's all that's necessary for what I'm suggesting.

How much of BB recentering is just decrease in torque and how much is the other leg doing to tug it that way?

That's kind of what I was asking above -- can we separate torque effects from non-torque effects? But I'm starting to think it doesn't matter. I'm not sure whether all the flex comes out of the frame at any phase of a crank revolution, now that we're looking at a number of different flex modes, but that's not necessary either -- what does come out still looks to me like it doesn't speed the bike up, but rather slows your legs down.

I think that was what we were originally trying to figure out?

[Mark McM]

Originally Posted by Kontact

And yet, the BB flex goes away before we pedal through 6 where you've shown that there is still plenty of down force on the pedal - down force that is much more in line with BB flex than it even is at 3 o'clock.

And what bit of magic effect do think does that? If there is differential in downward force at the pedals, there will be torque applied to the BB, and the frame will be flexed. All the pedal force vector diagrams show that there is still a downward force on the bottom pedal (which is greater than the force on the top pedal), so the frame will be flexed. That's just simple physics.

Originally Posted by Kontact

What is that, a royal fiat?


Did you read the article you linked to?

Yes, but I don't think you did. There is no royal fiat, just empirical evidence and numerical analysis. This page shows the strain energies from the different reaction forces. From the graph, the strain energy due to stays compressing (labeled the "Horizontal force" in the graph) is very small, just a few percent of the total strain energy in the frame.

Originally Posted by Kontact

If I take a flat strip of spring steel and twist the ends in opposite directions, would you not agree that the overall length of the strip gets shorter?

When I release that twist, does the strip not push itself back to full length?

You are really grasping. Yes, if the twist is very large, there will be very small change in length. But in this case, the twist is small enough that the change in length is essentially zero. This is covered in this text chapter on torsion reaction in structural members. Instead, the primary deflection is the rotation of the BB shell, resulting in vertical deflection at the pedals, and the primary strain energy is stored in this torsional twisting along an axis perpendicular to the crank torque.

[Mark McM]

Originally Posted by Kontact

This is one of those statements like "the man's feet don't push against the earth, the earth pushed up on his feet."

You can't argue that "helping the rider move their legs" is different than "helping the rider move the crank" in a system where net force never goes to zero. Until force goes to zero, the bike and the legs are all one mechanical linkage and you don't get to pick where the force is acting within it.

Nope, you're wrong again. Let's consider one crank revolution, starting with the leg at the top of the stroke: As the rider pushes down, the frame flexes. Some of the rider's energy goes directly into the drivetrain, and some goes into the frame in the form of strain energy. At the bottom of the pedal stroke, the rider still exerts a significant downward force on the pedal, so the frame is still flexed. As the rider continues the crank rotation, he has to expend energy to raise his leg. But as the frame un-flexes, the energy released by the frame helps the rider to raise his leg, so the rider's net energy is the same as if the frame had not flexed. But none of the strain energy went into the drivetrain - all the energy that went into the drivetrain occured during the downstroke. If a rider began from a standing start and only made one crank revolution, then it is very clear that the strain energy restoring the rider's leg position made no contribution to drive the bicycle. And it is made particularly clear if there is a time delay between the rider pushing all the way to the bottom of the stroke, and re-starting the revolution to bring their leg back up.